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8, , QUADRILATERALS, EXERCISE 8.1, , Q.1. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the, angles of the quadrilateral., Sol. Suppose the measures of four angles are 3x, 5x, 9x and 13x., ∴ 3x + 5x + 9x + 13x = 360°, [Angle sum property of a quadrilateral], ⇒, 30x = 360°, °, 360 = 12°, ⇒, x =, 30, ⇒, 3x = 3 × 12° = 36°, 5x = 5 × 12° = 60°, 9x = 9 × 12° = 108°, 13x = 13 × 12° = 156°, ∴ the angles of the quadrilateral are 36°, 60°, 108° and 156° Ans., Q.2. If the diagonals of a parallelogram are equal, then show that it is a, rectangle., Sol. Given : ABCD is a parallelogram in which AC = BD., To Prove : ABCD is a rectangle., Proof : In ∆ABC and ∆ABD, AB = AB, [Common], BC = AD, [Opposite sides of a parallelogram], AC = BD, [Given], ∴ ∆ABC ≅ ∆BAD, [SSS congruence], ∠ABC = ∠BAD, …(i), [CPCT], Since, ABCD is a parallelogram, thus,, ∠ABC + ∠BAD = 180°, …(ii), [Consecutive interior angles], ∠ABC + ∠ABC = 180°, ∴, 2∠ABC = 180°, [From (i) and (ii)], ⇒, ∠ABC = ∠BAD = 90°, This shows that ABCD is a parallelogram one of whose angle is 90°., Hence, ABCD is a rectangle., Proved., Q.3. Show that if the diagonals of a quadrilateral bisect each other at right, angles, then it is a rhombus., Sol. Given : A quadrilateral ABCD, in which diagonals AC and BD bisect each, other at right angles., To Prove : ABCD is a rhombus.

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Proof : In ∆AOB and ∆BOC, AO = OC, [Diagonals AC and BD bisect each other], ∠AOB = ∠COB, [Each = 90°], BO = BO, [Common], ∴ ∆AOB ≅ ∆BOC, [SAS congruence], AB = BC, …(i) [CPCT], Since, ABCD is a quadrilateral in which, AB = BC, [From (i)], Hence, ABCD is a rhombus., [∵ if the diagonals of a quadrilateral bisect each other, then it is a, parallelogram and opposite sides of a parallelogram are equal] Proved., Q.4. Show that the diagonals of a square are equal and bisect each other at, right angles., Sol. Given : ABCD is a square in which AC and BD are diagonals., To Prove : AC = BD and AC bisects BD at right angles, i.e. AC ⊥ BD., AO = OC, OB = OD, Proof : In ∆ABC and ∆BAD,, AB = AB, [Common], BC = AD, [Sides of a square], ∠ABC = ∠BAD = 90° [Angles of a square], ∴, ∆ABC ≅ ∆BAD, [SAS congruence], ⇒, AC = BD, [CPCT], Now in ∆AOB and ∆COD,, AB = DC, [Sides of a square], ∠AOB = ∠COD, [Vertically opposite angles], ∠OAB = ∠OCD, [Alternate angles], ∴, ∆AOB ≅ ∆COD, [AAS congruence], ∠AO = ∠OC, [CPCT], Similarly by taking ∆AOD and ∆BOC, we can show that OB = OD., In ∆ABC, ∠BAC + ∠BCA = 90°, [ ∠B = 90°], ⇒ 2∠BAC = 90°, [∠BAC = ∠BCA, as BC = AD], ⇒ ∠BCA = 45° or ∠BCO = 45°, Similarly ∠CBO = 45°, In ∆BCO., ∠BCO + ∠CBO + ∠BOC = 180°, ⇒ 90° + ∠BOC = 180°, ⇒ ∠BOC = 90°, ⇒ BO ⊥ OC ⇒ BO ⊥ AC, Hence, AC = BD, AC ⊥ BD, AO = OC and OB = OD., Proved., Q.5. Show that if the diagonals of a quadrilateral are equal and bisect each, other at right angles, then it is a square., Sol. Given : A quadrilateral ABCD, in which, diagonals AC and BD are equal and bisect, each other at right angles,, To Prove : ABCD is a square.

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Proof : Since ABCD is a quadrilateral whose diagonals bisect each other,, so it is a parallelogram. Also, its diagonals bisect each other at right, angles, therefore, ABCD is a rhombus., ⇒, AB = BC = CD = DA, [Sides of a rhombus], In ∆ABC and ∆BAD, we have, AB = AB, [Common], BC = AD, [Sides of a rhombus], AC = BD, [Given], ∴, ∆ABC ≅ ∆BAD, [SSS congruence], ∴, ∠ABC = ∠BAD, [CPCT], But, ∠ABC + ∠BAD = 180°, [Consecutive interior angles], ∠ABC = ∠BAD = 90°, ∠A = ∠B = ∠C = ∠D = 90°, [Opposite angles of a ||gm], ⇒ ABCD is a rhombus whose angles are of 90° each., Hence, ABCD is a square., Proved., Q.6. Diagonal AC of a parallelogram ABCD, bisects ∠A (see Fig.). Show that, (i) it bisects ∠C also,, (ii) ABCD is a rhombus., Given : A parallelogram ABCD, in which, diagonal AC bisects ∠A, i.e., ∠DAC = ∠BAC., To Prove : (i) Diagonal AC bisects, ∠C i.e., ∠DCA = ∠BCA, (ii) ABCD is a rhomhus., Proof :, (i), ∠DAC = ∠BCA, [Alternate angles], ∠BAC = ∠DCA, [Alternate angles], But, ∠DAC = ∠BAC, [Given], ∴, ∠BCA = ∠DCA, Hence, AC bisects ∠DCB, Or, AC bisects ∠C Proved., (ii) In ∆ABC and ∆CDA, AC = AC, [Common], ∠BAC = ∠DAC, [Given], and, ∠BCA = ∠DAC, [Proved above], ∴, ∆ABC ≅ ∆ADC, [ASA congruence], ∴, BC = DC, [CPCT], But AB = DC, [Given], ∴, AB = BC = DC = AD, Hence, ABCD is a rhombus Proved., [∵ opposite angles are equal], Q.7. ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠C and, diagonal BD bisects ∠B as well as ∠D., Sol. Given : ABCD is a rhombus, i.e.,, AB = BC = CD = DA., To Prove :, ∠DAC = ∠BAC,

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Sol. Given : In DABC and DDEF, AB = DE,, AB ||DE, BC = EF and BC || EF. Vertices A, B, and C are joined to vertices D, E and F., To Prove : (i), (ii), (iii), (iv), (v), (vi), Proof :, (i), , ABED is a parallelogram, BEFC is a parallelogram, AD || CF and AD = CF, ACFD is a parallelogram, AC = DF, ∆ABC ≅ ∆DEF, In quadrilateral ABED, we have, AB = DE and AB || DE., [Given], ⇒ ABED is a parallelogram., [One pair of opposite sides is parallel and equal], (ii) In quadrilateral BEFC, we have, BC = EF and BC || EF, [Given], ⇒ BEFC is a parallelogram., [One pair of opposite sides is parallel and equal], (iii) BE = CF and BE||BECF, [BEFC is parallelogram], AD = BE and AD||BE, [ABED is a parallelogram], ⇒ AD = CF and AD||CF, (iv) ACFD is a parallelogram., [One pair of opposite sides is parallel and equal], (v) AC = DF, [Opposite sides of parallelogram ACFD], (vi) In ∆ABC and ∆DEF, we have, AB = DE, [Given], BC = EF, [Given], AC = DF, [Proved above], ∴ ∆ABC ≅ ∆DEF, [SSS axiom] Proved., Q.12. ABCD is a trapezium in which AB, || CD and AD = BC (see Fig.)., Show that, (i) ∠ A = ∠ B, (ii) ∠C = ∠D, (iii) ∆ ABC ≅ ∆ BAD, (iv) diagonal AC = diagonal BD, Sol. Given : In trapezium ABCD, AB || CD and AD = BC., To Prove : (i) ∠A = ∠B, (ii) ∠C = ∠D, (iii) ∆ABC ≅ ∆BAD, (iv) diagonal AC = diagonal BD, Constructions : Join AC and BD. Extend AB and draw, a line through C parallel to DA meeting AB produced, at E.

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Proof :, , (i) Since, AB || DC, ⇒, AE || DC, …(i), and, AD || CE, …(ii), ⇒ ADCE is a parallelogram, ∠A + ∠E = 180°, , [Construction], [Opposite pairs of, sides are parallel, , …(iii), [Consecutive interior angles], ∠B + ∠CBE = 180°, …(iv), [Linear pair], AD = CE, …(v) [Opposite sides of a ||gm], AD = BC, …(vi), [Given], ⇒, BC = CE, [From (v) and (vi)], ⇒, ∠E = ∠CBE, …(vii), [Angles opposite to, equal sides], ∴ ∠B + ∠E = 180°, …(viii) [From (iv) and (vii), Now from (iii) and (viii) we have, ∠A + ∠E = ∠B + ∠E, ⇒, ∠A = ∠B Proved., (ii), ∠A + ∠D = 180°, ⎫, ⎬ [Consecutive interior angles], ∠B + ∠C = 180°, ⎭, ⇒ ∠A + ∠D = ∠B + ∠C, [∵ ∠A = ∠B], ⇒, ∠D = ∠C, Or, ∠C = ∠D Proved., (iii) In ∆ABC and ∆BAD, we have, AD = BC [Given], ∠A = ∠B [Proved], AB = CD, [Common], ∴ ∆ABC ≅ ∆BAD, [ASA congruence], (iv) diagonal AC = diagonal BD, [CPCT] Proved.

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QUADRILATERALS, , 8, , EXERCISE 8.2, Q.1. ABCD is a quadrilateral in which P, Q, R and S, are mid-points of the sides AB, BC, CD and DA, respectively. (see Fig.). AC is a diagonal. Show, that :, (i) SR || AC and SR =, , 1, AC, 2, , (ii) PQ = SR, (iii) PQRS is a parallelogram., Given : ABCD is a quadrilateral in which P, Q, R and S, are mid-points of AB, BC, CD and DA. AC is a diagonal., To Prove :, , Proof :, , (i) SR || AC and SR =, , 1, AC, 2, , (ii) PQ = SR, (iii) PQRS is a parallelogram, (i) In ∆ABC, P is the mid-point of AB and Q is the mid-point, of BC., , ∴ PQ || AC and PQ =, , 1, AC, 2, , …(1), , [Mid-point theorem], In ∆ADC, R is the mid-point of CD and S is the mid-point, of AD, ∴ SR || AC and SR =, , 1, AC, 2, , …(2), [Mid-point theorem], , (ii) From (1) and (2), we get, PQ || SR and PQ = SR, (iii) Now in quadrilateral PQRS, its one pair of opposite sides, PQ and SR is equal and parallel., ∴ PQRS is a parallelogram. Proved.

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Q.2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB,, BC, CD and DA respectively. Show that the quadrilateral PQRS is a, rectangle., Sol. Given : ABCD is a rhombus in which P, Q, R and S are mid points of, sides AB, BC, CD and DA respectively :, To Prove : PQRS is a rectangle., Construction : Join AC, PR and SQ., Proof : In ∆ABC, P is mid point of AB [Given], Q is mid point of BC [Given], 1, ⇒ PQ || AC and PQ = 2 AC …(i), Similarly, in ∆DAC,, , [Mid point theorem], , 1, SR || AC and SR = 2 AC, …(ii), From (i) and (ii), we have PQ||SR and PQ = SR, ⇒ PQRS is a parallelogram, [One pair of opposite sides is parallel and equal], Since ABQS is a parallelogram, ⇒ AB = SQ, [Opposite sides of a || gm], Similarly, since PBCR is a parallelogram., ⇒ BC = PR, Thus, SQ = PR, [AB = BC], Since SQ and PR are diagonals of parallelogram PQRS, which are equal., ⇒ PQRS is a rectangle. Proved., Q.3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC,, CD and DA respectively. Show that the quadrilataral PQRS is a rhombus., Sol. Given : A rectangle ABCD in which P, Q, R, S are, the mid-points of AB, BC, CD and DA respectively,, PQ, QR, RS and SP are joined., To Prove : PQRS is a rhombus., Construction : Join AC

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Proof : In ∆ABC, P and Q are the mid-points of the sides AB and BC., ∴ PQ || AC and PQ =, Similarly, in ∆ADC,, , 1, AC, 2, , …(i), , [Mid point theorem], , 1, AC, …(ii), 2, From (i) and (ii), we get, PQ || SR and PQ = SR, …(iii), Now in quadrilateral PQRS, its one pair of opposite sides PQ and SR is, parallel and equal, [From (iii)], ∴PQRS is a parallelogram., Now, AD = BC, …(iv), [Opposite sides of a rectangle ABCD], , SR || AC and SR =, , 1, 1, AD =, BC, 2, 2, ⇒, AS = BQ, In ∆APS and ∆BPQ, AP = BP, [∵ P is the mid-point of AB], AS = BQ, [Proved above], ∠PAS = ∠PBQ, [Each = 90°], ∆APS ≅ ∆BPQ, [SAS axiom], ∴, PS = PQ, …(v), From (iii) and (v), we have, PQRS is a rhombus Proved., Q.4. ABCD is a trapezium in which, AB || DC, BD is a diagonal and E is the, mid-point of AD. A line is drawn through E, parallel to AB intersecting BC at F (see, Fig.). Show that F is the mid-point of BC., Sol. Given : A trapezium ABCD with, AB || DC, E is the mid-point of AD and EF, || AB., O, To Prove : F is the mid-point of BC., Proof : AB || DC and EF || AB, ⇒ AB, EF and DC are parallel., Intercepts made by parallel lines AB, EF and DC on transversal AD are, equal., ∴ Intercepts made by those parallel lines on transversal BC are also, equal., i.e., BF = FC, ⇒ F is the mid-point of BC., Q.5. In a parallelogram ABCD, E and F are the, mid-points of sides AB and CD respectively, (see Fig.). Show that the line segments AF, and EC trisect the diagonal BD., , ∴

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Sol. Given : A parallelogram ABCD, in which, E and F are mid-points of sides AB and DC, respectively., To Prove : DP = PQ = QB, Proof : Since E and F are mid-points of AB and DC respectively., ⇒ AE =, , 1, 1, AB and CF =, DC, 2, 2, , But, AB = DC and AB || DC, , …(i), …(ii), [Opposite sides of a parallelogram], , ∴ AE = CF and AE || CF., ⇒ AECF is a parallelogram., [One pair of opposite sides is parallel and equal], In ∆BAP,, E is the mid-point of AB, EQ || AP, ⇒ Q is mid-point of PB, [Converse of mid-point theorem], ⇒, PQ = QB, …(iii), Similarly, in ∆DQC,, P is the mid-point of DQ, DP = PQ, …(iv), From (iii) and (iv), we have, DP = PQ = QB, or line segments AF and EC trisect the diagonal BD. Proved., Q.6. Show that the line segments joining the mid-points of the opposite sides of, a quadrilateral bisect each other., Sol. Given : ABCD is a quadrilateral in which EG and, FH are the line segments joining the mid-points of, opposite sides., To Prove : EG and FH bisect each other., Construction : Join EF, FG, GH, HE and AC., Proof : In ∆ABC, E and F are mid-points of AB and BC respectively., ∴ EF =, , 1, AC and EF || AC, 2, , …(i), , In ∆ADC, H and G are mid-points of AD and CD respectively., ∴ HG =, , 1, AC and HG || AC, 2, , …(ii), , From (i) and (ii), we get, EF = HG and EF || HG, ∴ EFGH is a parallelogram., [∵ a quadrilateral is a parallelogram if its, one pair of opposite sides is equal and parallel], Now, EG and FH are diagonals of the parallelogram EFGH., ∴ EG and FH bisect each other., [Diagonal of a parallelogram bisect each other] Proved.

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Q.7. ABC is a triangle right angled at C. A line through the mid-point M of, hypotenuse AB and parallel to BC intersects AC at D. Show that, (i) D is the mid-point of AC., (ii) MD ⊥ AC, 1, AB, 2, Sol. Given : A triangle ABC, in which ∠C = 90° and M is the mid-point of AB, and BC || DM., To Prove : (i) D is the mid-point of AC, [Given], (ii) DM ⊥ BC, , (iii) CM = MA =, , (iii) CM = MA =, , 1, AB, 2, , Construction : Join CM., Proof : (i) In ∆ABC,, M is the mid-point of AB., [Given], BC || DM, [Given], D is the mid-point of AC, [Converse of mid-point theorem] Proved., (ii), ∠ADM = ∠ACB, [∵ Coresponding angles], But, ∠ACB = 90°, [Given], ∴, ∠ADM = 90°, But ∠ADM + ∠CDM = 180°, [Linear pair], ∴, ∠CDM = 90°, Hence, MD ⊥ AC Proved., (iii) AD = DC …(1), [ ∵ D is the mid-point of AC], Now, in ∆ADM and ∆CMD, we have, ∠ADM = ∠CDM, [Each = 90°], AD = DC, [From (1)], DM = DM, [Common], ∴, ∆ADM ≅ ∆CMD, [SAS congruence], ⇒, CM = MA, …(2) [CPCT], Since M is mid-point of AB,, ∴, , MA =, , 1, AB, 2, , Hence, CM = MA =, , 1, AB, 2, , …(3), Proved. [From (2) and (3)]